The correct option is C (−1,0)
f(x)=(a−2)(x−[x])2+2(x−[x])+a2=0
Let t=x−[x]={x}
⇒t∈[0,1)
(a−2)t2+2t+a2=0
It is given that the given equation does not have any integer root.
So t≠0 (∵∀ x∈Z,x=[x]⇒t=0)
∴t∈(0,1)
Case 1: a−2≠0
f(t)=(a−2)t2+2t+a2=0
For f(x) to have exactly one root in [2,3), f(t) should have exactly one root in the interval (0,1).
∴f(0)f(1)<0
⇒a2(a2+a)<0
⇒a∈(−1,0)
Case 2: a−2=0
Then 2t+a2=0 which is not possible as LHS>0
∴a∈(−1,0)